//定义一个100元素的集合,包含A-Z List<String> list = new LinkedList<>(); for (int i =0;i<100;i++){ list.add(String.valueOf((char)('A'+Math.random()*('Z'-'A'+1)))); } System.out.println(list); //统计集合重复元素出现次数,并且去重返回hashmap Map<String, Long> map = l
//统计数组中出现次数超过一半的数字 #include <stdio.h> int Find(int *arr, int len) { int num = 0; //当前数字 int times = 0; //当前数字出现的次数 int i = 0; for (i = 0; i<len; i++) { if (times == 0) { num = arr[i]; times = 1; } else if (arr[i] == num) times++; else times--; }
1.方法一 var arr = [1, 2, 3, 1, 2, 4]; function arrayCnt(arr) { var newArr = []; for(var i = 0; i < arr.length; i++) { if(newArr.indexOf(arr[i]) == -1) { newArr.push(arr[i]) } } var newarr2 = new Array(newArr.length); for(var t = 0; t < newarr2.length;
import java.util.HashMap; public class map1 { public static void main(String[] args) { String[] array = {"a","b","a","b","c","a","b","c","b"}; HashMap<String, Int